On the size of a non-empty family of non-empty sets such that every set in the family has a proper subset also in the family Let $ F$ be a non-empty family of non-empty sets such that for every set $A \in F$ , $\exists $ a proper subset $B \subset A$ such that $B \in F$ . I can prove that every set in such a family $F$ is infinite but I'm having trouble to determine whether the family $F$ itself is infinite or not ; I have an intuition that the family must also be infinite but am not able to prove ( or disprove) it ; please help . Thanks

$F$ is non-empty. Then there is a set $A_1 \in F$. Now as per the conditions there is a propoer subset $A_2 \subset A_1$ such that $A_2 \in F$. You can continue to induct that there is a sequence $A_1, A_2, ...A_n, ..$ such that each of them are elements of $F$.

Or proceed by contradiction. Suppose there are but a finite number of elements in $F$. Say, $F = \\{B_1, B_2, .., B_n\\}$. Now there must be a set $C$ in the finite collection $F$ such that there is no other set in $F$ which is a proper subset of $C$. The absence of such a set $C$ immediately entails that $F$ is infinite. But then since $C$ is in $F$ there must be a proper subset of it which is also in $F$ leading to a contradiction. Hence $F$ must be an infinite collection.